My lecturer was talking today (in the context of probability, more specifically Kolmogorov's axioms) about the additive property of functions, namely that:

$$f(x+y) = f(x) + f(y)$$

I've been trying to find what functions satisfy this. Intuition says that, for functions over $\mathbb{R}$, the only functions should be of the form $f(x) = ax$ for some real a. Unfortunately I've only shown this is true when the domain of the function is the rational multiples of a given real number.

My question is if it is possible to extend this result (that $f(x) = ax$ given additivity) to the real numbers, possibly without assuming the continuity of f. It seems to me that additivity introduces so many constrains on a function that nothing but the trivial case would be able to sneak through. The following is a summary of my thoughts to date, though they're obviously long and not 'compulsory reading'. :)

When x is rational - Preliminary Investigation

It is not hard to see that:

$$f(x + x + x) = 3f(x)$$

and more generally, for $a\in \mathbb{N}$, $$f(ax) = af(x)$$ It is not too hard to prove (well, it took half a bus trip ... ) that this also applies first for $a\in \mathbb{Z}$ and then for $a\in \mathbb{Q}$, (for the latter you just need to consider $a=m/n$ and then note that:

$$f\Big(\frac{m}{n}x\Big)=mf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot nf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot f\Big(n\frac{x}{n}\Big)=\frac{m}{n}\cdot f(x)$$

The reason this little equation is cool is that we can set $x = 1$ and get:

$$f(a)=a\cdot f(1)$$

which is equivalent to what was expected intuitively, namely (after changing $a$ to $y$ and $f(1)$ to $a$)

$$f(y) = a\cdot y$$

as long as y is rational

$y$ is a rational multiple of a real number

But we can do a bit better than that. If we substitute in $x = \sqrt{2}$ or any other real number in $f(ax) = af(x)$ (which we know for rational $a$), you can conduct the exact same argument above and show that, for instance

$$f(y) = \Big(\frac{f(\sqrt{2})}{\sqrt{2}}\Big)\cdot y=a\cdot y$$

Whenever $y = \frac{m}{n}\sqrt{2}$ i.e. whenever $y$ is a rational multiple of $\sqrt{2}$. Note however, that the value of the coefficient $a$ (i.e. the slope of the line) is apparently completely unrelated to the value taken in the case where $y$ is purely rational.

What I'm actually asking

We still haven't shown that $$f(x) = ax$$ for all $x \in \mathbb{R}$, as the slope of the line may change depending on what real number we are taking rational multiples of. As far as I've shown now, we might have $f(x) = x$ when $x$ is rational, $f(x) = 3x$ when $x$ is a rational multiple of $\sqrt{2}$, etc.

I still feel that $f(x) = ax$ for all $x \in \mathbb{R}$. One reason for thinking this comes from noting that $$f(2) = f(2-\sqrt{2})+f(\sqrt{2})$$

$2$, $2-\sqrt{2}$ and $\sqrt{2}$ are not rational multiples of each other, however the equation above gives a restraint on the slopes of the lines formed by their rational multiples (which we'll call $a_1, a_2$ and $a_3$ for the slopes on the rational multiples of $2, 2-\sqrt{2}$ and $\sqrt{2}$ respectively). We have $2a_1 = (2-\sqrt{2}) a_2 + \sqrt{2} a_3$

There's so many constraints here - all the rational multipes have the same coefficient, whenever 2 (or more) numbers which aren't rational multiples of each other are added together we get another constraint on their coefficients. The trivial solution is just that$$f(x) = ax$$

over $x \in \mathbb{R}$ and I really struggle to see how any other solution could possible squeeze through all these constraints.

Is there an additive function on $\mathbb{R}$ not of the form $f(x) = ax$?


Yea. See the Wikipedia article on Cauchy's Functional Equation.

  • 3
    $\begingroup$ It’s probably worth pointing out that they require the axiom of choice and therefore cannot be constructed explicitly. $\endgroup$ – Brian M. Scott Mar 1 '12 at 11:27
  • 1
    $\begingroup$ They're also discontinuous, so if you have a standing assumption that $f$ is continuous then you're correct. And your proof is almost complete as well, you can use a limiting argument on $f\left(\frac{m}{n}x\right)=\frac{m}{n}f(1)$ to finish it. $\endgroup$ – mdp Mar 1 '12 at 11:58

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